Positive Solutions for Fractional Differential Equations from Real Estate Asset Securitization via New Fixed Point Theorem

Abstract

We study a fractional differential equation dynamics model arising from the analysis of real estate asset securitization by using the generalized fixed point theorem for weakly contractive mappings in partially ordered sets. Based on the analysis for the existence and uniqueness of the solution and scientific numerical calculation of the solution, in further study, some optimization schemes for traditional risk control process will be obtained, and then the main results of this paper can be applied to the forefront of research of real estate asset securitization.

1. Introduction

Real estate asset securitization is a kind of important financial derivatives in the world. In the international capital market, by combining with different development pattern of the financial industry and real estate industry, real estate asset securitization has become a class of financial products with rapid development and great vitality. Recently, by SWOT analysis method, one has found that many mathematical models arising from real estate asset securitization can be interpreted by fractional-order differential or difference equations, under suitable initial conditions or boundary conditions, the existence and uniqueness of solution of the fractional-order mechanical model are important and useful. Especially, by examining the numerical simulation and analysis of solution, one can undertake macroscopical analysis and comparative research for real estate securitization process advantages and disadvantages, and find real estate asset securitization may exist problems and risks, and then one can put forward to optimize the views on traditional risk control process. In recent years, fractional-order models have been proved to be more accurate than integer order models; that is, there are more degrees of freedom in the fractional-order models, see [1, 2]. For recent works about fractional-order models, we refer the reader to [3–9] and the references therein.

In this paper, we discuss the existence and uniqueness of positive solutions for the following fractional differential equation with nonlocal Riemann-Stieltjes integral condition arising from the real estate asset securitization −𝒟_𝐭^𝛼𝑥(𝑡)=𝑓𝑡,𝑥(𝑡),−𝒟_𝐭^𝛽𝑥(𝑡)(),𝒟,𝑡∈0,1_𝐭𝛽𝑥(0)=𝒟_𝐭^𝛽+1𝑥(0)=0,𝒟_𝐭^𝛽𝑥(1)=¹₀𝒟_𝐭^𝛽𝑥(𝑠)𝑑𝐴(𝑠), (1.1) where 2<𝛼≤3, 0<𝛽<1, and 𝛼−𝛽>2, 𝒟_𝐭 is the standard Riemann-Liouville derivative.

In (1.1), ∫¹₀𝒟_𝐭^𝛽𝑥(𝑠)𝑑𝐴(𝑠) denotes the Riemann-Stieltjes integral, and 𝐴 is a function of bounded variation, which implies 𝑑𝐴 can be a signed measure. Thus the BVP (1.1) reveal that positive or negative values of the linear functionals ∫¹₀𝑥(𝑠)𝑑𝐴(𝑠) are viable in some cases. But if 𝐴(𝑠)=𝑠 or 𝑑𝐴(𝑠)=ℎ(𝑠)𝑑𝑠, the BVP become a multipoint boundary value problems or the integral boundary value problems, some kind of positivity on the functionals ∫¹₀𝑥(𝑠)𝑑𝐴(𝑠) is often required, for example, in ∫¹₀𝑥(𝑠)𝑑𝐴(𝑠)=∑^𝑚−2_𝑖=1𝜇_𝑖𝑢(𝜂_𝑖), 𝜇_𝑖>0 is often required. Thus the BVP (1.1) include more generalized boundary value conditions. For a detailed description of multipoint and integral boundary conditions on fractional differential equation, we refer the reader to some recent papers (see [10–15]).

Note that the problem (1.1) has been considered by Wang and Li [16], and the authors obtained the existence of one positive solution for (1.1) by using Krasnoselskii’s fixed point theorem on a cone under semipositone case, but the uniqueness of the solution is not treated.

The paper is organized as follows. In Section 2, we extend and improve some fixed point theorems for weakly contractive mappings in partially ordered sets by Harjani and Sadarangani [17], we omit many redundant conditions of Theorems 2 and  3 in [17] and obtain better results than those of [17]. In Section 3, the existence and uniqueness of a positive solution for the problem (1.1) are obtained by using a fixed point theorem in partially ordered sets. An example is also given to illuminate the application of the main result.

2. Preliminaries and Lemmas

Definition 2.1 (see [18, 19]). The Riemann-Liouville fractional integral of order 𝛼>0 of a function 𝑥∶(0,+∞)→ℝ is given by 𝐼^𝛼𝑥(𝑡)=1Γ(𝛼)^𝑡₀(𝑡−𝑠)^𝛼−1𝑥(𝑠)𝑑𝑠 (2.1) provided that the right-hand side is pointwise defined on (0,+∞).

Definition 2.2 (see [18, 19]). The Riemann-Liouville fractional derivative of order 𝛼>0 of a function 𝑥∶(0,+∞)→ℝ is given by 𝒟_𝐭^𝛼𝑥(𝑡)=1Γ()𝑑𝑛−𝛼𝑑𝑡^𝑛𝑡₀(𝑡−𝑠)^{𝑛−𝛼−1}𝑥(𝑠)𝑑𝑠, (2.2) where 𝑛=[𝛼]+1, [𝛼] denotes the integer part of number 𝛼, provided that the right-hand side is pointwise defined on (0,+∞).

Remark 2.3. If 𝑥,𝑦∶(0,+∞)→ℝ with order 𝛼>0, then 𝒟_𝐭^𝛼𝑥(𝑡)+𝑦(𝑡)=𝒟_𝐭^𝛼𝑥(𝑡)+𝒟_𝐭^𝛼𝑦(𝑡). (2.3)

Proposition 2.4 (see [18, 19]). (1) If 𝑥∈𝐿¹(0,1), 𝜈>𝜎>0, then 𝐼^𝜈𝐼^𝜎𝑥(𝑡)=𝐼^𝜈+𝜎𝑥(𝑡),𝒟_𝐭^𝜎𝐼^𝜈𝑥(𝑡)=𝐼^𝜈−𝜎𝑥(𝑡),𝒟_𝐭^𝜎𝐼^𝜎𝑥(𝑡)=𝑥(𝑡). (2.4)
(2) If 𝜈>0, 𝜎>0, then 𝒟_𝐭^𝜈𝑡^𝜎−1=Γ(𝜎)Γ()𝑡𝜎−𝜈^{𝜎−𝜈−1}. (2.5)

Proposition 2.5 (see [18, 19]). Let 𝛼>0, and 𝑓(𝑥) is integrable, then 𝐼^𝛼𝒟_𝐭^𝛼𝑓(𝑥)=𝑓(𝑥)+𝑐₁𝑥^𝛼−1+𝑐₂𝑥^𝛼−2+⋯+𝑐_𝑛𝑥^𝛼−𝑛, (2.6) where 𝑐_𝑖∈ℝ (𝑖=1,2,…,𝑛), 𝑛 is the smallest integer greater than or equal to 𝛼.

Let 𝑥(𝑡)=𝐼^𝛽𝑦(𝑡), 𝑦(𝑡)∈𝐶[0,1], by standard discuss, we easily reduce the BVP (1.1) to the following modified problems: −𝒟_𝐭^𝛼−𝛽𝑦(𝑡)=𝑓𝑡,𝐼^𝛽𝑦(𝑡)(𝑡),𝑦(0),−𝑦=𝑦^′(0)=0,𝑦(1)=¹₀𝑦(𝑠)𝑑𝐴(𝑠), (2.7) and the BVP (2.7) is equivalent to the BVP (1.1).

Lemma 2.6 (see [3]). Given 𝑦∈𝐿¹(0,1), then the problem 𝒟_𝐭^𝛼−𝛽𝑥(𝑡)(𝑡)𝑥(0)+𝑦=0,0<𝑡<1,=𝑥^′(0)=0,𝑥(1)=0, (2.8) has the unique solution 𝑥(𝑡)=¹₀𝐺(𝑡,𝑠)𝑦(𝑠)𝑑𝑠, (2.9) where 𝐺(𝑡,𝑠) is given by 𝐺()=1𝑡,𝑠Γ[𝑡(𝛼−𝛽1−𝑠)]^{𝛼−𝛽−1}[𝑡(,0≤𝑡≤𝑠≤1,1−𝑠)]^{𝛼−𝛽−1}−(𝑡−𝑠)^{𝛼−𝛽−1},0≤𝑠≤𝑡≤1, (2.10) which is the Green function of the BVP (2.8).

Lemma 2.7 (see [3]). For any 𝑡,𝑠∈[0,1], 𝐺(𝑡,𝑠) satisfies 𝑡^{𝛼−𝛽−1}(1−𝑡)𝑠(1−𝑠)^{𝛼−𝛽−1}Γ()≤𝑠()𝛼−𝛽≤𝐺𝑡,𝑠1−𝑠^{𝛼−𝛽−1}Γ.𝛼−𝛽−1 (2.11)

By Lemma 2.6, the unique solution of the problem 𝒟_𝐭^𝛼𝑥(𝑡)𝑥(0)=0,0<𝑡<1,=𝑥^′(0)=0,𝑥(1)=1, (2.12) is 𝑡^{𝛼−𝛽−1}. Let 𝒞=¹₀𝑡^{𝛼−𝛽−1}𝑑𝐴(𝑡), (2.13) and define 𝒢_𝐴(𝑠)=¹₀𝐺(𝑡,𝑠)𝑑𝐴(𝑡), (2.14) as in [4], we can get that the Green function for the nonlocal BVP (2.7) is given by 𝐻()=𝑡𝑡,𝑠^{𝛼−𝛽−1}𝒢1−𝒞^𝐴(𝑠)+𝐺(𝑡,𝑠). (2.15)

Throughout the paper, we always assume the following holds. (H0)𝐴 is a increasing function of bounded variation such that 𝒢_𝐴(𝑠)≥0 for 𝑠∈[0,1] and 0≤𝒞<1, where 𝒞 is defined by (2.13).

Lemma 2.8. Let 1<𝛼−𝛽≤2 and (H0) hold, then 𝐻(𝑡,𝑠) satisfies ()≤10≤𝐻𝑡,𝑠()Γ.1−𝒞𝛼−𝛽−1 (2.16)

Proof. By (2.11) and that 𝐴(𝑡) is a increasing function of bounded variation, we have 𝒢_𝐴(𝑠)=¹₀𝐺(𝑡,𝑠)𝑑𝐴(𝑡)≤¹₀𝑡^{𝛼−𝛽−1}Γ(𝑡)=𝒞𝛼−𝛽−1𝑑𝐴Γ.𝛼−𝛽−1 (2.17) Consequently, 𝐻()≤𝒢𝑡,𝑠^𝐴(𝑠)+𝑠()1−𝒞1−𝑠^{𝛼−𝛽−1}Γ≤1𝛼−𝛽−1()Γ.1−𝒞𝛼−𝛽−1 (2.18)

Now, we present a result about the fixed point theorems which is improvement of [17].

Theorem 2.9. Let (𝑋,≥) be a partially ordered set and suppose that there exists a metric 𝑑 in 𝑋 such that (𝑋,𝑑) is a complete metric space. Assume that 𝑋 satisfies the following condition: if 𝑥_𝑛 is a nondecreasing sequence in 𝑋 such that 𝑥_𝑛→𝑥 then 𝑥_𝑛≤𝑥 for all 𝑛∈ℕ. Let 𝑇∶𝑋→𝑋 be a nondecreasing mapping, and there exists a constant 𝜆∈(0,1) such that 𝑑𝑇𝑥,𝑇𝑦≤𝜆𝑑𝑥,𝑦−𝜓𝜆𝑑𝑥,𝑦,for𝑥≥𝑦, (2.19) where 𝜓∶[0,+∞)→[0,+∞) is a nondecreasing function. If there exists 𝑥∈𝑋 with 𝑥₀≤𝑇𝑥₀, then 𝑇 has a fixed point.

Proof. If 𝑇(𝑥₀)=𝑥₀, then the proof is finished. Suppose that 𝑥₀<𝑇(𝑥₀). Since 𝑇 is a nondecreasing mapping, we obtain by induction that 𝑥₀<𝑇𝑥₀≤𝑇²𝑥₀≤𝑇³𝑥₀≤⋯≤𝑇^𝑛𝑥₀≤⋯. (2.20) Put 𝑥_𝑛+1=𝑇^𝑛(𝑥₀)=𝑇𝑥_𝑛. Then for each integer 𝑛≥1, from (2.20), the elements 𝑥_𝑛 and 𝑥_𝑛+1 are comparable, we get 𝑑𝑥_𝑛+1,𝑥_𝑛=𝑑𝑇𝑥_𝑛,𝑇𝑥_𝑛−1≤𝜆𝑑𝑥_𝑛,𝑥_𝑛−1−𝜓𝜆𝑑𝑥_𝑛,𝑥_𝑛−1≤𝜆𝑑𝑥_𝑛,𝑥_𝑛−1. (2.21)
If there exists 𝑛₀∈ℕ such that 𝑑(𝑥_𝑛0,𝑥_𝑛0−1)=0 then 𝑥_𝑛0=𝑇^𝑛0−1(𝑥₀)=𝑥_𝑛0−1 and 𝑥_𝑛0−1 is a fixed point and the proof is finished. In other case, suppose that 𝑑(𝑥_𝑛0,𝑥_𝑛0−1)≠0 for all 𝑛∈ℕ. Then by (2.21), we have 𝑑𝑥_𝑛+1,𝑥_𝑛≤𝜆𝑑𝑥_𝑛,𝑥_𝑛−1≤⋯≤𝜆^{𝑛−𝑛0+1}𝑑𝑥_𝑛0,𝑥_𝑛0−1→0,as𝑛→+∞, (2.22) that is, 𝜌_𝑛=𝑑𝑥_𝑛+1,𝑥_𝑛→0,as𝑛→+∞. (2.23) Similar to [17], 𝑥_𝑛 is a Cauchy sequence and then there exists 𝑧∈𝑋 such that lim_𝑛→+∞𝑥_𝑛=𝑧 since (𝑋,𝑑) is a complete metric space.
We claim that 𝑇(𝑧)=𝑧. In fact, 𝑑(𝑇𝑧,𝑧)≤𝑑𝑇𝑧,𝑇𝑥_𝑛+𝑑𝑇𝑥_𝑛,𝑧≤𝜆𝑑𝑧,𝑥_𝑛−𝜓𝜆𝑑𝑧,𝑥_𝑛+𝑑𝑥_𝑛+1,𝑧≤𝜆𝑑𝑧,𝑥_𝑛+𝑑𝑥_𝑛+1,𝑧 (2.24) and taking limit as 𝑛→+∞, 𝑑(𝑇𝑧,𝑧)≤0, this proves that 𝑑(𝑇𝑧,𝑧)=0, consequently, 𝑇(𝑧)=𝑧.

Remark 2.10. In Theorem 2.9, We do not require that 𝜓 is continuous and 𝜓 is positive in (0,+∞), 𝜓(0)=0 and lim_𝑡→+∞𝜓(𝑡)=+∞. This in essence improve and generalize the corresponding results of Theorem 2 and Theorem 3 in paper [17].
If we consider that (𝑋,≤) satisfies the following condition: for𝑥,𝑦∈𝑋,thereexists𝑧∈𝑋whichiscomparableto𝑥and𝑦, (2.25) then we have the following theorem, see [17].

Theorem 2.11. Adding condition (2.25) to the hypotheses of Theorem 2.9, one obtains uniqueness of the fixed point of 𝑇.

In our considerations, we will work in the Banach space 𝐶[0,1]={𝑥∶[0,1]→ℝiscontinuous} with the standard norm ||𝑥||=max_0≤𝑡≤1|𝑥(𝑡)|.

Note that this space can be equipped with a partial order given by 𝑥,𝑦∈𝐶[0,1],𝑥≤𝑦⟺𝑥(𝑡)≤𝑦(𝑡),for𝑡∈[0,1]. (2.26)

In [20, 21], it is proved that (𝐶[0,1],≤) with the classic metric given by 𝑑𝑥,𝑦=max_0≤𝑡≤1𝑥(𝑡)−𝑦(𝑡) (2.27) satisfies the following condition.

If 𝑥_𝑛 is a nondecreasing sequence in 𝑋 such that 𝑥_𝑛→𝑥 then 𝑥_𝑛≤𝑥 for all 𝑛∈ℕ. Moreover, for 𝑥,𝑦∈𝐶[0,1], as the function max{𝑥,𝑦} is continuous in [0, 1], and (𝐶[0,1],≤) satisfies condition (2.25).

3. Main Results

Define the class of function 𝒜, if 𝜙∈𝒜, then 𝜙∶[0,+∞)→[0,+∞) is nondecreasing, and 𝜓(𝑥)=𝑥−𝜙(𝑥) satisfies 𝜓∶[0,+∞)→[0,+∞) is nondecreasing.

The standard functions 𝜙∈𝒜, for example, 𝜙(𝑥)=arctan𝑥, 𝜙(𝑥)=ln(1+𝑥) and 𝜙(𝑥)=𝑥/(1+𝑥), and so forth.

Theorem 3.1. Suppose (H0) holds, 𝑓∶[0,1]×[0,+∞)×(−∞,0,]→[0,+∞) is continuous, and for any fixed 𝑡∈[0,1], 𝑓(𝑡,𝑥,𝑦) is nondecreasing in 𝑥 on [0,+∞) and nonincreasing in 𝑦 on (−∞,0]; moreover, there exist two positive constants 𝜌₁, 𝜌₂ that satisfy 𝜌₁+𝜌₂≤Γ𝛼−𝛽−1(1−𝒞), (3.1) and there exist a function 𝜙∈𝒜 and constants 0<𝜃₁<1/(Γ(𝛽+1)), 0<𝜃₂<1 such that 𝑓𝑡,𝑥₁,𝑦₁−𝑓𝑡,𝑥₂,𝑦₂≤𝜌₁𝜙𝜃₁𝑥₁−𝑥₂+𝜌₂𝜙𝜃₂𝑦₂−𝑦₁ (3.2) for 𝑥_𝑖,𝑦_𝑖∈[0,+∞), 𝑖=1,2 with 𝑥₁≥𝑥₂, 𝑦₁≤𝑦₂ and 𝑡∈[0,1].
Then problem (1.1) has a unique nonnegative solution.

Proof. Consider the cone 𝑃=𝑦∈𝐶[0,1]∶𝑦(𝑡)≥0. (3.3) Note that, as 𝑃 is a closed set of 𝐶[0,1], 𝑃 is a complete metric space.
Now, for 𝑦∈𝑃, we define the operator 𝑇 by 𝑇𝑦(𝑡)=¹₀𝐻(𝑡,𝑠)𝑓𝑠,𝐼^𝛽𝑦(𝑠),−𝑦(𝑠)𝑑𝑠. (3.4) Then from the assumption on 𝑓 and Lemma 2.8, we have 𝑇(𝑃)⊂𝑃. (3.5)
In what follows, we check that hypotheses in Theorems 2.9 and 2.11 are satisfied.
Firstly, the operator 𝑇 is nondecreasing since, by hypothesis, for any 𝑢≥𝑣(𝑇𝑢)(𝑡)=¹₀𝐻(𝑡,𝑠)𝑓𝑠,𝐼^𝛽𝑢(𝑠),−𝑢(𝑠)𝑑𝑠≥¹₀𝐻(𝑡,𝑠)𝑓𝑠,𝐼^𝛽𝑣(𝑠),−𝑣(𝑠)𝑑𝑠=(𝑇𝑣)(𝑡). (3.6)
Besides, for 𝑢≥𝑣𝑑(𝑇𝑢,𝑇𝑣)=max_[]𝑡∈0,1|𝑢(𝑡)−𝑣(𝑡)|=max_𝑡∈[0,1]¹₀𝐻(𝑡,𝑠)𝑓𝑠,𝐼^𝛽𝑢(𝑠),−𝑢(𝑠)−𝑓𝑠,𝐼^𝛽𝑣(𝑠),−𝑣(𝑠)𝑑𝑠≤max_𝑡∈[0,1]¹₀𝐻(𝑡,𝑠)𝜌₁𝜙𝜃₁𝐼^𝛽𝑢(𝑠)−𝐼^𝛽𝑣(𝑠)+𝜌₂𝜙𝜃₂(𝑢(𝑠)−𝑣(𝑠))𝑑𝑠. (3.7) Since 𝐼^𝛽𝑢(𝑠)−𝐼^𝛽𝑣(𝑠)=1Γ𝛽^𝑡₀(𝑡−𝑠)^𝛽−1(𝑢(𝑠)(𝑠≤𝑑()−𝑣))𝑑𝑠𝑢,𝑣Γ𝛽^𝑡₀(𝑡−𝑠)^𝛽−1𝑑()𝑑𝑠≤𝑢,𝑣Γ,𝛽+1 (3.8) and 𝜙 is nondecreasing, for 𝑢≥𝑣, we have 𝜌₁𝜙𝜃₁𝐼^𝛽𝑢(𝑠)−𝐼^𝛽𝑣(𝑠)+𝜌₂𝜙𝜃₂(𝑢(𝑠)−𝑣(𝑠))≤𝜌₁𝜙𝜃₁Γ𝑑()𝛽+1𝑢,𝑣+𝜌₂𝜙𝜃₂𝑑()≤𝜌𝑢,𝑣₁+𝜌₂𝜙(𝜃𝑑(𝑢,𝑣)), (3.9) where 𝜃𝜃=max¹Γ𝛽+1,𝜃₂∈(0,1). (3.10) Note that 𝜙∈𝒜, then 𝜓(𝑥)=𝑥−𝜙(𝑥), and 𝜓∶[0,+∞)→[0,+∞) is nondecreasing. Thus by Lemma 2.8, for 𝑢≥𝑣, we have 𝑑(𝑇𝑢,𝑇𝑣)≤𝜌₁+𝜌₂𝜙(𝜃𝑑(𝑢,𝑣))max_[]𝑡∈0,1¹₀𝐻()((()−()((()((.𝑡,𝑠𝑑𝑠≤𝜙𝜃𝑑𝑢,𝑣))=𝜃𝑑𝑢,𝑣𝜃𝑑𝑢,𝑣−𝜙𝜃𝑑𝑢,𝑣))=𝜃𝑑𝑢,𝑣−𝜓𝜃𝑑𝑢,𝑣)) (3.11)
Finally, taking into account that the zero function, 0≤𝑇0, by Theorem 2.11, problem (1.1) has a unique nonnegative solution.

Theorem 3.2. If the assumptions of Theorem 3.1 are satisfied, and there exists 𝑡₀∈[0,1] such that 𝑓(𝑡₀,0,0)≠0, then the unique solution of (1.1) is positive. (Positive solution means a solution satisfying 𝑥(𝑡)>0 for 𝑡∈(0,1).)

Proof. By Theorem 3.1, the problem (1.1) has a unique nonnegative solution, which also is positive.
In fact, otherwise, there exists 0<𝑡^∗<1 such that 𝑥(𝑡^∗)=0, and 𝑥𝑡^∗=¹₀𝐻𝑡^∗,𝑠𝑓𝑠,𝐼^𝛽𝑥(𝑠),−𝑥(𝑠)𝑑𝑠=0. (3.12) Then 0=𝑥𝑡^∗=¹₀𝐻𝑡^∗,𝑠𝑓𝑠,𝐼^𝛽𝑥(𝑠),−𝑥(𝑠)𝑑𝑠≥¹₀𝐻𝑡^∗,𝑠𝑓(𝑠,0,0)𝑑𝑠≥0. (3.13) Consequently, ¹₀𝐻𝑡^∗,𝑠𝑓(𝑠,0,0)𝑑𝑠=0, (3.14) this yields 𝐻𝑡^∗,𝑠𝑓(𝑠,0,0)=0,a.e. (3.15) Note that 𝐻(𝑡^∗,𝑠)>0,𝑠∈(0,1), then we have 𝑓(𝑠,0,0)=0,a.e. (3.16)
But on the other hand, since 𝑓(𝑡₀,0,0)≠0, 𝑡₀∈[0,1], we have 𝑓(𝑡₀,0,0)>0, by the continuity of 𝑓, we can find a set Ω⊂[0,1] satisfying 𝑡₀∈Ω and the Lebesgue measure 𝜇(Ω)>0 such that 𝑓(𝑡,0)>0 for any 𝑡∈Ω. This contradicts (3.16). Therefore, 𝑥(𝑡)>0, that is, 𝑥(𝑡) is positive solution of (1.1).

Example 3.3. Consider the following boundary value problem with fractional order 𝛼=21/8: −𝒟^21/8𝑥(𝑡)=𝑒^𝑡+𝑥(𝑡)((𝑡+𝒟1001+𝑥))^1/8𝑥(𝑡)3001+𝒟^1/8𝑥(𝑡)𝒟,0<𝑡<1,_1/8𝑥(0)=𝒟^9/8𝑥(0)=0,𝒟^1/8𝑥(1)=¹₀𝒟^1/8𝑥(𝑠)𝑑𝐴(𝑠), (3.17) where 𝐴(𝑡)=10,𝑡∈0,4,11,𝑡∈4,34,32,𝑡∈4,1. (3.18) Thus the BVP (3.17) becomes the 4-Point BVP with coefficients −𝒟^21/8𝑥(𝑡)=𝑒^𝑡+𝑥(𝑡)((𝑡+𝒟1001+𝑥))^1/8𝑥(𝑡)3001+𝒟^1/8𝑥(𝑡)𝒟,0<𝑡<1,_1/8𝑥(0)=𝒟^9/8𝑥(0)=0,𝒟^1/8𝑥(1)=𝒟^1/8𝑥14+𝒟^1/8𝑥34. (3.19) Then the BVP (3.17) has a unique positive solution.

Proof. Obviously, 𝛼=21/8, 𝛽=1/8, and 0≤𝒞=¹₀𝑡^3/2(𝑡)=1𝑑𝐴4^3/2+34^3/2≈0.7745<1,Γ𝛼−𝛽−1(1−𝒞)=0.2014. (3.20)
On the other hand, we have 𝐺()=𝐺𝑡,𝑠¹()=[𝑡(𝑡,𝑠1−𝑠)]^3/2Γ()𝐺5/2,0≤𝑡≤𝑠≤1,₂()=[𝑡(𝑡,𝑠1−𝑠)]^3/2−(𝑡−𝑠)^3/2Γ()𝒢5/2,0≤𝑠≤𝑡≤1,_𝐴(𝑠)=𝐺₂14,𝑠+2𝐺₂341,𝑠,0≤𝑠<4,𝐺₁14,𝑠+2𝐺₂34,1,𝑠43≤𝑠<4,𝐺₁14,𝑠+2𝐺₁34,3,𝑠4≤𝑠≤1. (3.21) Thus 𝒢_𝐴(𝑠)≥0, 0≤𝒞<1 and 𝐴(𝑠) is increasing. So (H0) holds.
Take 𝑓(𝑡,𝑢,𝑣)=𝑒^𝑡+𝑢()−𝑣1001+𝑢(),()∈[]×[)×(].3001−𝑣𝑡,𝑢,𝑣0,10,+∞−∞,0 (3.22) Then 𝑓𝑡,𝑥(𝑡),−𝒟_𝐭^1/8𝑥(𝑡)=𝑒^𝑡+𝑥(𝑡)((𝑡+𝒟1001+𝑥))^1/8𝑥(𝑡)3001+𝒟^1/8𝑥(𝑡), (3.23) and for any 𝑥₁≥𝑥₂, 𝑦₁≤𝑦₂, 𝑓𝑡,𝑥₁,𝑦₁−𝑓𝑡,𝑥₂,𝑦₂=𝑥₁1001+𝑥₁−𝑦₁3001−𝑦₁−𝑥₂1001+𝑥₂+𝑦₂3001−𝑦₂=𝑥₁−𝑥₂1001+𝑥₁1+𝑥₂+𝑦₂−𝑦₁3001−𝑦₁1−𝑦₂≤𝑥₁−𝑥₂1001+𝑥₁−𝑥₂+𝑦₂−𝑦₁3001+𝑦₂−𝑦₁≤1×𝑥100¹−𝑥₂1+1/2𝑥₁−𝑥₂+1×𝑦300²−𝑦₁1+1/3𝑦₂−𝑦₁=1×(𝑥501/2)₁−𝑥₂1+(1/2)𝑥₁−𝑥₂+1×(𝑦1001/3)₂−𝑦₁1+(1/3)𝑦₂−𝑦₁=1𝜙1502𝑥₁−𝑥₂+1𝜙11003𝑦₂−𝑦₁, (3.24) where 𝜙(𝑥)=𝑥1+𝑥,𝜌₁+𝜌₂=1+150𝜃100=0.03<0.2014,₁=12<1Γ=1𝛽+1Γ98=1.0619,𝜃₂=13. (3.25) Thus 𝜙∈𝒜 and all of condition of Theorem 3.1 are satisfied.
On the other hand, 𝑓(𝑡,0,0)=𝑒^𝑡≠0 for any 𝑡∈[0,1], by Theorem 3.2, the BVP (3.17) has a unique positive solution.

In the end, we claim this process and conclusion of paper can be applied to the forefront of research of the real estate asset securitization.